Finite Math Examples

$x - z + 3 y = 4$ , $z = 3 y$ , $y - x = 5 z$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Move $- z$ .

$x + 3 y - z = 4$

$z = 3 y$

$y - x = 5 z$

Step 1.2

Subtract $3 y$ from both sides of the equation.

$x + 3 y - z = 4$

$z - 3 y = 0$

$y - x = 5 z$

Step 1.3

Reorder $z$ and $- 3 y$ .

$x + 3 y - z = 4$

$- 3 y + z = 0$

$y - x = 5 z$

Step 1.4

Subtract $5 z$ from both sides of the equation.

$x + 3 y - z = 4$

$- 3 y + z = 0$

$y - x - 5 z = 0$

Step 1.5

Reorder $y$ and $- x$ .

$x + 3 y - z = 4$

$- 3 y + z = 0$

$- x + y - 5 z = 0$

$x + 3 y - z = 4$

$- 3 y + z = 0$

$- x + y - 5 z = 0$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & - 3 & 1 & 0 \\ - 1 & 1 & - 5 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation $R_{3} = R_{3} + R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.1.1

Perform the row operation $R_{3} = R_{3} + R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & - 3 & 1 & 0 \\ - 1 + 1 \cdot 1 & 1 + 1 \cdot 3 & - 5 - 1 & 0 + 1 \cdot 4 \end{array}]$

Step 3.1.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & - 3 & 1 & 0 \\ 0 & 4 & - 6 & 4 \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 1 & - \frac{1}{3} \cdot 0 \\ 0 & 4 & - 6 & 4 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 4 & - 6 & 4 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & - 6 - 4 (- \frac{1}{3}) & 4 - 4 \cdot 0 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 0 & - \frac{14}{3} & 4 \end{array}]$

Step 3.4

Multiply each element of $R_{3}$ by $- \frac{3}{14}$ to make the entry at $3, 3$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{3}$ by $- \frac{3}{14}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ - \frac{3}{14} \cdot 0 & - \frac{3}{14} \cdot 0 & - \frac{3}{14} (- \frac{14}{3}) & - \frac{3}{14} \cdot 4 \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 3.5

Perform the row operation $R_{2} = R_{2} + \frac{1}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{2} = R_{2} + \frac{1}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 + \frac{1}{3} \cdot 0 & 1 + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & 0 + \frac{1}{3} (- \frac{6}{7}) \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 3.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 3.6

Perform the row operation $R_{1} = R_{1} + R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} + R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + 0 & 3 + 0 & - 1 + 1 \cdot 1 & 4 - \frac{6}{7} \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 3 & 0 & \frac{22}{7} \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 3.7

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 0 - 3 \cdot 0 & \frac{22}{7} - 3 (- \frac{2}{7}) \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 4

Use the result matrix to declare the final solution to the system of equations.

$x = 4$

$y = - \frac{2}{7}$

$z = - \frac{6}{7}$

Step 5

The solution is the set of ordered pairs that make the system true.

$(4, - \frac{2}{7}, - \frac{6}{7})$